Prove that $A_5$ does not have a subgroup of order 15, 20, nor 30 (without using simple group and Sylow's Theorem)

Question

Show that $A_5$ does not have a subgroup of order 15, 20, nor 30.

I looked up other posts such as the one below, but they used some things we don't cover in the course.

$A_5$ has no subgroup of order 15 and 20

Instead, we did a similar example in class:

Show that $A_4$ has no subgroups of order 6.

Suppose $H \preceq A_4$ with $|H| = 6$. Then $[A_4: H] = 2$ and so $H \unlhd A_4$. Therefore, $(aH)^2 = H \forall a \in A_4$. This means $a^2 \in H \forall a \in A_4$. We can compute to show that $A_4$ has 9 elements of the form $a^2$, and by above they must all be in $H$. But this is a contradiction since $|H| = 6$. Hence no such $H$ exist.

But the problem with trying to apply a similar proof to my question is that, since $[A_5: H]$ = 4, we cannot say that $H \unlhd A_5$. Hence, we cannot say $(aH)^4 = H$ since we cannot form the factor group $A_5/H$. If that is the case, how else can I solve the question without using simple group and Sylow's Theorem.

Answer 1

So, we've pretty much narrowed it down to the subgroup of order $20$.

To tackle that, note that if $H\le A_5$ with $\mid H\mid=20$, then let $A_5$ act on the set of $3$ left cosets $A_5/H$. We get a homomorphism from $A_5$ into $S_3$, which is non-trivial: it's kernel $\cap_{x\in G}xHx^{-1}\subset H$.

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Briefly, the homomorphism is defined by letting $\varphi (g)$ be the element of $S_3$ corresponding to the permutation of the cosets brought about by applying $g$. That's let $\bar{g_1},\bar{g_2},\bar{g_3}$ be the cosets. Multiplying by $g$ on the left permutes them. We get $\bar{g_1}\to \bar {gg_1},\,\bar {g_2}\to\bar{gg_2}$ and $\bar {g_3}\to\bar{gg_3}$. It's easy to see that $g$ takes different cosets to different cosets. So we really do have an element of $S_3$.

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Now to continue note that $A_5=\langle(12)(34),(135)\rangle$. And $(12)(34)(135)$ has order $5$. Then $e=\varphi ((12)(34)(135))=\varphi ((12)(34))\varphi ((135))\implies \varphi ((12)(34))=\varphi ((135))=e$, because the orders divide $2$ and $3$ (and are the same). So we've reached a contradiction.

Answer 2

A very low-tech approach.

Since $30$ is an integer of the form $4k+2$ a subgroup with $30$ elements would automatically admit a subgroup with $15$ elements. By Cauchy's theorem a subgroup of $A_5$ with $15$ elements would contain a $5$-cycle $\varphi$ and a $3$-cycle $\gamma$. However $\varphi\gamma$ or $\varphi\gamma^2$ necessarily contains a transposition in its cycle decomposition, which makes $|\langle \varphi,\gamma\rangle|$ even.

About subgroups of order $20$: in this case we would have a $5$-cycle $\varphi$ (which we may assume to be $(12345)$) and a double transposition $\delta$. Up to conjugation we may freely assume that $1$ is a fixed point of $\delta$, and restrict our attention to three cases:

1. $\delta=(23)(45)$. In this case $\delta\varphi$ has order $3$ and we are doomed.
2. $\delta=(24)(35)$. In this case the product between $\delta$ and its conjugated $(14)(35)$ has order $3$. We are doomed.
3. $\delta=(25)(34)$. Here we may take the identity, all the double transpositions given by $\varphi^k \delta \varphi^{-k}$ and the powers of $\varphi$ to actually have a subgroup of $A_5$ with order $10$. That is not enough. Since we cannot take any extra double transposition, we need to take other $10$ $5$-cycles from the $5$-cycles in $A_5$ which are not powers of $\varphi$. At that point we would have more than half the $5$-cycles in $A_5$, and the span would be $A_5$. Or: by conjugating $\delta$ via one of these new $5$-cycles we would get a double transposition of the "bad" kind, being doomed again.

Answer 3

Here is another proof that $A_5$ does not contain subgroups of order $20$.

Let $H<A_5$ and $|H|=20$.

1. Not all cycles of length $5$ are contained in $H$ (since $A_5$ has $24$ cycles of length $5$).

2. If $a$ is a cycle of length $5$ and $a\notin H$, then $a^2,a^3\notin H$ and the cosets $H,Ha,Ha^2$ are pairwise distinct, so $$ G=H\cup Ha\cup Ha^2. $$

3. Which of the three cosets contains $a^3$? Contradiction.

Adding. I find this helpful.

If $H$ is a subgroup of $A_5$, then either $|H|\leq12$ or $H=A_5$.

Proof without using Sylow's theorem, the simplicity of the group $A_n$, and the notion of a group action. But the proof uses the Lagrange theorem.

1. There exists a cycle $a$ of length $5$ which is not contained in the subgroup $H$. Then $a^2,a^3,a^4\notin H$ and so the cosets $H,Ha,Ha^2,Ha^3,Ha^4$ are pairwise different. Hence $|G:H|\geq5$ and $|H|\leq12$.

2. Let all cycles of length $5$ be contained in $H$. Since $$ (i_1i_2i_3i_4i_5)(i_1i_2i_5i_4i_3)=(i_1i_3i_2), $$ it follows that all cycles of length $3$ are contained in $H$. There are $24$ cycles of length $5$ and $20$ cycles of length $3$. So $|H|>44\Rightarrow|H|=60\Rightarrow H=A_5$.